Answer:
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Explanation:
Answer: The energy system related to your question is missing attached below is the energy system.
answer:
a) Work done = Net heat transfer
Q1 - Q2 + Q + W = 0
b) rate of work input ( W ) = 6.88 kW
Explanation:
Assuming CPair = 1.005 KJ/Kg/K
<u>Write the First law balance around the system and rate of work input to the system</u>
First law balance ( thermodynamics ) :
Work done = Net heat transfer
Q1 - Q2 + Q + W = 0 ---- ( 1 )
rate of work input into the system
W = Q2 - Q1 - Q -------- ( 2 )
where : Q2 = mCp T = 1.65 * 1.005 * 293 = 485.86 Kw
Q2 = mCp T = 1.65 * 1.005 * 308 = 510.74 Kw
Q = 18 Kw
Insert values into equation 2 above
W = 6.88 Kw
Answer:
There is 0.466 KW required to operate this air-conditioning system
Explanation:
<u>Step 1:</u> Data given
Heat transfer rate of the house = Ql = 755 kJ/min
House temperature = Th = 24°C = 24 +273 = 297 Kelvin
Outdoor temperature = To = 35 °C = 35 + 273 = 308 Kelvin
<u>Step 2: </u> Calculate the coefficient of performance o reversed carnot air-conditioner working between the specified temperature limits.
COPr,c = 1 / ((To/Th) - 1)
COPr,c = 1 /(( 308/297) - 1)
COPr,c = 1/ 0.037
COPr,c = 27
<u>Step 3:</u> The power input cna be given as followed:
Wnet,in = Ql / COPr,max
Wnet, in = 755 / 27
Wnet,in = 27.963 kJ/min
Win = 27.963 * 1 KW/60kJ/min = 0.466 KW
There is 0.466 KW required to operate this air-conditioning system
Answer:
Schematics
Explanation:
A schematic is a detailed structured diagram or drawing. It employs illustrations to help the viewer understand detailed information on the machine or object being described. Its main aim is not to help the observer know what the object looks like physically. It is rather aimed at helping the viewer know how the machine works. This is achieved by only including key and important details to the drawing.
It is most times used in the blueprint and user guides of machines and gadgets used in the home to help users know how these things work so that they can do little fixings should there be such needs.
Answer:
The elevation at the high point of the road is 12186.5 in ft.
Explanation:
The automobile weight is 2500 lbf.
The automobile increases its gravitational potential energy in
. It means the mobile has increased its elevation.
The initial elevation is of 5183 ft.
The first step is to convert Btu of potential energy to adequate units to work with data previously presented.
British Thermal Unit -
Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:
Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.
is the final elevation and
is the initial elevation.
Replacing W in the Ep equation
Finally, the next step is to replace the variables of the problem.
The elevation at the high point of the road is 12186.5 in ft.